Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2

Question 1.
Write the correct number in the given boxes from the following A.P.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2 1
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2 2

Question 2.
Decide whether following sequence is an A.P., if so find the 20th term of the progression.
-12, -5, 2, 9,16, 23,30,…
Solution:
i. The given sequence is
-12, -5,2, 9, 16, 23,30,…
Here, t1 = -12, t2 = -5, t3 = 2, t4 = 9
∴ t2 – t1 – 5 – (-12) – 5 + 12 = 7
t3 – t2 = 2 – (-5) = 2 + 5 = 7
∴ t4 – t3 – 9 – 2 = 7
∴ t2 – t1 = t3 – t2 = … = 7 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P.

ii. tn = a + (n – 1)d
∴ t20 = -12 + (20 – 1)7 …[∵a = -12, d = 7]
= -12 + 19 × 7
= -12 + 133
∴ t20 = 121
∴ 20th term of the given A.P. is 121.

Question 3.
Given Arithmetic Progression is 12, 16, 20, 24, … Find the 24th term of this progression.
Solution:
The given A.P. is 12, 16, 20, 24,…
Here, a = 12, d = 16 – 12 = 4 Since,
tn = a + (n – 1)d
∴ t24 = 12 + (24 – 1)4
= 12 + 23 × 4
= 12 + 92
∴ t24 = 104
∴ 24th term of the given A.P. is 104.

Question 4.
Find the 19th term of the following A.P. 7,13,19,25…..
Solution:
The given A.P. is 7, 13, 19, 25,…
Here, a = 7, d = 13 – 7 = 6
Since, tn = a + (n – 1)d
∴ t19 = 7 + (19 – 1)6
= 7 + 18 × 6
= 7 + 108
∴ t19 = 115
∴ 19th term of the given A.P. is 115.

Question 5.
Find the 27th term of the following A.P. 9,4,-1,-6,-11,…
Solution:
The given A.P. is 9, 4, -1, -6, -11,…
Here, a = 9, d = 4- 9 = -5
Since, tn = a + (n – 1)d
∴ t27 = 9 + (27 – 1)(-5)
= 9 + 26 × (-5)
= 9 – 130
∴ t27 = -121
∴ 27th term of the given A.P. is -121.

Question 6.
Find how many three digit natural numbers are divisible by 5.
Solution:
The three digit natural numbers divisible by
5 are 100, 105, 110, …,995
The above sequence is an A.P.
∴ a = 100, d = 105 – 100 = 5
Let the number of terms in the A.P. be n.
Then, tn = 995
Since, tn = a + (n – 1)d
∴ 995 = 100 +(n – 1)5
∴ 995 – 100 = (n – 1)5
∴ 895 = (n – 1)5
∴ n – 1 = \(\frac { 895 }{ 5 } \)
∴ n – 1 = 179
∴ n = 179 + 1 = 180
∴ There are 180 three digit natural numbers which are divisible by 5.

Question 7.
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
Solution:
Bor an A.P., let a be the first term and d be the common difference,
t11 = 16, t21 = 29 …[Given]
tn = a + (n – 1)d
∴ t11, = a + (11 – 1)d
∴ 16 = a + 10d
i.e. a + 10d = 16 …(i)
Also, t21 = a + (21 – 1)d
∴ 29 = a + 20d
i.e. a + 20d = 29 …(ii)
Subtracting equation (i) from (ii), we get a
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2 3

Question 8.
8. 11, 8, 5, 2, … In this A.P. which term is number-151?
Solution:
The given A.P. is 11, 8, 5, 2,…
Here, a = 11, d = 8 – 11 = -3
Let the nth term of the given A.P. be -151.
Then, tn = – 151
Since, tn = a + (n – 1)d
∴ -151= 11 + (n – 1)(-3)
∴ -151 – 11 =(n – 1)(-3)
∴ -162 = (n – 1)(-3)
∴ n – 1 = \(\frac { -162 }{ -3 } \)
∴ n – 1 = 54
∴ n = 54 + 1 = 55
∴ 55th term of the given A.P. is -151.

Question 9.
In the natural numbers from 10 to 250, how many are divisible by 4?
Solution:
The natural numbers from 10 to 250 divisible
by 4 are 12, 16, 20, …,248
The above sequence is an A.P.
∴ a = 12, d = 16 – 12 = 4
Let the number of terms in the A.P. be n.
Then, tn = 248
Since, tn = a + (n – 1)d
∴ 248 = 12 + (n – 1)4
∴ 248 – 12 = (n – 1)4
∴ 236 = (n – 1)4
∴ n – 1 = \(\frac { 236 }{ 4 } \)
∴ n – 1 = 59
∴ n = 59 + 1 = 60
∴ There are 60 natural numbers from 10 to 250 which are divisible by 4.

Question 10.
In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
Solution:
For an A.P., let a be the first term and d be the common difference.
According to the given condition,
t17 = t10 + 7
∴ a + (17 – 1)d = a + (10 – 1)d + 7 …[∵ tn = a + (n – 1)d]
∴ a + 16d = a + 9d + 7
∴ a + 16d – a – 9d = 7
∴ 7d = 7
∴ d = \(\frac { 7 }{ 7 } \) = 1
∴ The common difference is 1.

Question 1.
Kabir’s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, “it seems like Kabir’s height grows in Arithmetic Progression”. Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years. (Textbook pg. no. 63)
Solution:
Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm
Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P.
Here, a = 70, d = 80 – 70 = 10
We have to find height of Kabir at the age of 15years i.e. t15.
Now, tn = a + (n – 1)d
∴ t15 = 70 + (15 – 1)10
= 70 + 14 × 10 = 70 + 140
∴ t15 = 210
∴ The height of Kabir at the age of 15 years will be 210 cm.

Question 2.
Is 5, 8, 11, 14, …. an A.P.? If so then what will be the 100th term? Check whether 92 is in this A.P.? Is number 61 in this A.P.? (Textbook pg. no, 62)
Solution:
i. The given sequence is
5, 8,11,14,…
Here, t1 = 5, t2 = 8, t3 = 11, t4 = 14
∴ t2 – t1 = 8 – 5 = 3
t3 – t2 = 11 – 8 = 3
t4 – t3 = 14 – 11 = 3
∴ t2 – t1 = t3 – t2 = t4 – t3 = 3 = d = constant
The difference between two consecutive terms is constant
∴ The given sequence is an A.P.

ii. tn = a + (n – 1)d
∴ t100 = 5 + (100 – 1)3 …[∵ a = 5, d = 3]
= 5 + 99 × 3
= 5 + 297
∴ t100 = 302
∴ 100th term of the given A.P. is 302.

iii. To check whether 92 is in given A.P., let tn = 92
∴ tn = a + (n – 1)d
∴ 92 = 5 + (n – 1)3
∴ 92 = 5 + 3n – 3
∴ 92 = 2 + 3n
∴ 90 = 3n
∴ n = \(\frac { 90 }{ 3 } \) = 30
∴ 92 is the 30th term of given A.P.

iv. To check whether 61 is in given A.P., let tn = 61
61 = 5 + (n – 1)3
∴ 61 = 5 + 3n – 3
∴ 61 = 2 + 3n
∴ 61 – 2 = 3n
∴ 59 = 3n
∴ n = \(\frac { 59 }{ 3 } \)
But, n is natural number 59
∴ n ≠ \(\frac { 59 }{ 3 } \)
∴ 61 is not in given A.P.

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2

Question 1.
For each of the following experiments write sample space ‘S’ and number of sample Point n(S)
i. One coin and one die are thrown simultaneously.
ii. Two digit numbers are formed using digits 2,3 and 5 without repeating a digit.
Solution:
i. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) =12
ii. Sample space,
S = {23,25,32, 35, 52, 53}
∴ n(S) = 6

Question 2.
The arrow is rotated and it stops randomly on the disc. Find out on which colour it may stop.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 1
Solution:
There are total six colours on the disc.
Sample space,
S = {Red, Orange, Yellow, Blue, Green, Purple}
∴ n(S) = 6
∴ Arrow may stop on any one of the six colours.

Question 3.
In the month of March 2019, find the days on which the date is a multiple of 5. (see the given page of the calendar).
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 2
Solution:
Dates which are multiple of 5:
5,10, 15,20,25,30
∴ S = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}
∴ n(S) = 6
∴ The days on which the date will be a multiple of 5 are Tuesday, Sunday, Friday, Wednesday, Monday and Saturday.

Question 4.
Form a ‘Road safety committee’ of two, from 2 boys (B1 B2) and 2 girls (G1, G2). Complete the following activity to write the sample space.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 3

Question 1.
Sample Space

  • The set of all possible outcomes of a random experiment is called sample space.
  • It is denoted by ‘S’ or ‘Ω’ (omega).
  • Each element of a sample space is called a sample point.
  • The number of elements in the set S is denoted by n(S).
  • If n(S) is finite, then the sample space is called a finite sample space.

Some examples of finite sample space. (Textbook pg. no, 117)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 4 Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 5

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1

Question 1.
How many possibilities are there in each of the following?
i. Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation. Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.
ii. Any day of a week is to be selected randomly.
iii. Select one card from the pack of 52 cards.
iv. One number from 10 to 20 is written on each card. Select one card randomly.
Solution:
i. Here, 8 sites of Maharashtra are given.
∴ There are 8 possibilities in a random experiment of visiting a site out of 8 sites in Maharashtra.

ii. There are 7 days in a week.
∴ There are 7 possibilities in a random experiment of selecting a day of the week.

iii. There are 52 cards in a pack of cards.
∴ There are 52 possibilities in a random experiment of selecting one card from the pack of 52 cards.

iv. There are 11 cards numbered from 10 to 20.
∴ There are 11 possibilities in a random experiment of selecting one card from the given set of cards.

Question 1.
In which of the following experiments possibility of expected outcome is more? (Textbook pg, no. 116)
i. Getting 1 on the upper face when a die is thrown.
ii. Getting head by tossing a coin.
Solution:
i. On a die there are 6 numbers.
∴ There are 6 possibilities of getting any one number from 1 to 6 on the upper face i.e. \(\frac { 1 }{ 6 } \) is the possibility.

ii. There are two possibilities (H or T) on tossing a coin i.e. \(\frac { 1 }{ 2 } \) possibility.
∴ In the second experiment, the possibility of expected outcome is more.

Question 2.
Throw a die, once. What are the different possibilities of getting dots on the upper face? (Textbook pg. no. 114)
Answer:
There are six different possibilities of getting dots on the upper face. They are
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1

Question 1.
Which of the following sequences are A.P.? If they are A.P. find the common difference.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 1
Solution:
i. The given sequence is 2, 4, 6, 8,…
Here, t1 = 2, t2 = 4, t3 = 6, t4 = 8
∴ t2 – t1 = 4 – 2 = 2
t3 – t2 = 6 – 4 = 2
t4 – t3 = 8 – 6 = 2
∴ t2 – t1 =  t3 – t2 = … = 2 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 2.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 2
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 3
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = \(\frac { 1 }{ 2 } \).

iii. The given sequence is -10, -6, -2, 2,…
Here, t1 = -10, t2 = – 6, t3 = -2, t4 = 2
∴ t2 – t1 = -6 – (-10) = -6 + 10 = 4
t3 – t2 = -2 -(-6) = -2 + 6 = 4
t4 – t3 = 2 – (-2) = 2 + 2 = 4
∴ t2 – t1 = t3 – t2 = … = 4 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 4.

iv. The given sequence is 0.3, 0.33, 0.333,…
Here, t1 = 0.3, t2 = 0.33, t3 = 0.333
∴ t2 -t1 = 0.33 – 0.3 = 0.03
t3 – t2 = 0.333 – 0.33 = 0.003
∴ t2 – t1 ≠ t3 – t2
The difference between two consecutive terms is not constant.
∴ The given sequence is not an A.P.

v. The given sequence is 0, -4, -8, -12,…
Here, t1 = 0, t2 = -4, t3 = -8, t4 = -12
∴ t2 – t1 = -4 – 0 = -4
t3 – t2 = -8 – (-4) = -8 + 4 = -4
t4 – t3 = -12 – (-8) = -12 + 8 = -4
∴ t2 – t1 = t3 – t2 = … = —4 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = -4.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 4
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 0.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 5
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = √2.

viii. The given sequence is 127, 132, 137,…
Here, t1 = 127, t2 = 132, t3 = 137
∴ t2 – t1 = 132 – 127 = 5
t3 – t2 = 137 – 132 = 5
∴ t2 – t1 = t3 – t2 = … = 5 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 5.

Question 2.
Write an A.P. whose first term is a and common difference is d in each of the following.
i. a = 10, d = 5
ii. a = -3, d = 0
iii. a = -7, d = \(\frac { 1 }{ 2 } \)
iv. a = -1.25, d = 3
v. a = 6, d = -3
vi. a = -19, d = -4
Solution:
i. a = 10, d = 5 …[Given]
∴ t1 = a = 10
t2 = t1 + d = 10 + 5 = 15
t3 = t2 + d = 15 + 5 = 20
t4 = t3 + d = 20 + 5 = 25
∴ The required A.P. is 10,15, 20, 25,…

ii. a = -3, d = 0 …[Given]
∴ t1 = a = -3
t2 = t1 + d = -3 + 0 = -3
t3 = t2 + d = -3 + 0 = -3
t4 = t3 + d = -3 + 0 = -3
∴ The required A.P. is -3, -3, -3, -3,…

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 6
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 7
∴ The required A.P. is -7, – 6.5, – 6, – 5.5,

iv. a = -1.25, d = 3 …[Given]
t1 = a = -1.25
t2 = t1 + d = – 1.25 + 3 = 1.75
t3 = t2 + d = 1.75 + 3 = .4.75
t4 = t3 + d = 4.75 + 3 = 7.75
∴ The required A.P. is -1.25, 1.75, 4.75, 7.75,…

v. a = 6, d = -3 …[Given]
∴ t1 = a = 6
t2 = t1 + d = 6 – 3 = 3
t3 = t2 + d = 3 – 3 = 0
t4 = t3 + d = 0- 3 = -3
∴ The required A.P. is 6, 3, 0, -3,…

vi. a = -19, d = -4 …[Given]
t1 = a = -19
t2 = t1 + d = -19 – 4 = -23
t3 = t2 + d = -23 – 4 = -27
t4 = t3 + d = -27 – 4 = -31
∴ The required A.P. is -19, -23, -27, -31,…

Question 3.
Find the first term and common difference for each of the A.P.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 8
Solution:
i. The given A.P. is 5, 1,-3,-7,…
Here, t1 = 5, t2 = 1
∴ a = t1 = 5 and
d = t2 – t1 = 1 – 5 = -4
∴ first term (a) = 5,
common difference (d) = -4

ii. The given A.P. is 0.6, 0.9, 1.2, 1.5,…
Here, t1 = 0.6, t2 = 0.9
∴ a = t1 = 0.6 and
d = t2 – t1 = 0.9 – 0.6 = 0.3
∴ first term (a) = 0.6,
common difference (d) = 0.3

iii. The given A.P. is 127, 135, 143, 151,…
Here, t1 = 127, t2 = 135
∴ a = t1 = 127 and
d = t2 – t1 = 135 – 127 = 8
∴ first term (a) = 127,
common difference (d) = 8

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 9

Question 1.
Complete the given pattern. Look at the pattern of the numbers. Try to find a rule to obtain the next number from its preceding number. Write the next numbers. (Textbook pg, no. 55 and 56)
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 10
Every pattern is formed by adding a circle in horizontal and vertical rows to the preceding pattern.
∴ The sequence for the above pattern is 1,3, 5, 7, 9,11,13,15,17,….
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 11
Every pattern is formed by adding 2 triangles horizontally and 1 triangle vertically to the preceding pattern.
∴ The sequence for the above pattern is 5,8,11,14,17,20,23,…

Question 2.
Some sequences are given below. Show the positions of the terms by t1, t2, t3,…
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 12

Question 3.
Some sequences are given below. Check whether there is any rule among the terms. Find the similarity between two sequences. To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably. (Textbook pg. no. 56 and 57)
i. .1,4,7,10,13,…
ii. 6,12,18,24,…
iii. 3,3,3,3,…
iv. 4, 16, 64,…
v. -1, -1.5, -2, -2.5,…
vi. 13, 23, 33, 43
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 13
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 14
The similarity in the sequences i., ii., iii. and v. is that the next term is obtained by adding a particular fixed number to the previous term.

Note : A Geometric Progression is a sequence in which the ratio of any two consecutive terms is a constant,
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 15
Sequence iv. is a geometric progression.

Question 4.
Write one example of finite and infinite A.P. each. (Textbook pg. no. 59)
Answer:
Finite A.P.:
Even natural numbers from 4 to 50:
4, 6, 8, ………………. 50.
Infinite A. P.:
Positive multiples of 5:
5, 10, 15, ……………..

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3

Question 1.
Choose the correct alternative answer for each of the following sub questions.

i. The sequence – 10,- 6,- 2, 2, …
(A) is an A.P. Reason d = – 16
(B) is an A.P. Reason d = 4
(C) is an A.P. Reason d = – 4
(D) is not an A.P.
Answer:
(B)

ii. First four terms of an A.P. are …, whose first term is -2 and common difference is -2.
(A) -2, 0, 2, 4
(B) -2, 4,- 8, 16
(C) -2, -4, -6, -8
(D) -2,-4, -8, -16
Answer:
(C)

iii. What is the sum of the first 30 natural numbers?
(A) 464
(B) 465
(C) 462
(D) 461
Answer:
(B)

iv. For an given A.P. t7 = 4, d = – 4, then a = ………
(A) 6
(B) 7
(C) 20
(D) 28
Answer:
(D)

v. For an given A.P. a = 3.5, d = 0, n = 101, then tn = ….
(A) 0
(B) 3.5
(c) 103.5
(D) 104.5
Answer:
(B)

vi. In an A.P. first two terms are – 3, 4, then 21st term is ….
(A) -143
(B) 143
(C) 137
(D) 17
Answer:
(C)

vii. If for any A.P. d = 5, then t18 – t13 = ….
(A) 5
(B) 20
(C) 25
(D) 30
Answer:
(C)

viii. Sum of first five multiples of 3 is …
(A) 45
(B) 55
(C) 15
(D) 75
Answer:
(A)

ix. 15, 10, 5, … In this A.P. sum of first 10 terms is…
(A) -75
(B) -125
(C) 75
(D) 125
Answer:
(A)

x. In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is 399, then n = ….
(A) 42
(B) 38
(C) 21
(D) 19
Answer:
(B)

Hints:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 1

Question 2.
Find the fourth term from the end in an
A.P.: -11, -8, -5, …, 49.
Solution:
The given A.P. is
-11,-8,-5, ……. 49
Reversing the A.P., we get 49, …, -5, -8, -11
Here, a = 49, d = -11 -(-8) = -11 + 8 = -3
Since, tn = a + (n – 1)d
∴ t4 = 49 + (4 – 1)(-3)
= 49 + (3) (-3)
= 49 – 9
= 40
∴ Fourth term from the end in the given A.P. is 40.
[Note: If an AY. is reversed, then the resulting sequence is also an A.P.]

Question 3.
In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P.
Solution:
For an A.P., let a be the first term and d be the common difference.
t10 = 46, t5 + t7 = 52 …[Given]
Since, tn = a + (n – 1)d
∴ t10 = a + (10 – 1)d
∴ 46 = a + 9d
i. e. a + 9d = 46 …(i)
Also, t5 + t7 = 52
∴ a + (5 – 1)d + a + (7 – 1)d = 52
∴ a + 4d + a + 6d = 52
∴ 2a + 10d = 52
∴ 2 (a + 5d) = 52
∴ a + 5d = \(\frac { 52 }{ 2 } \)
∴ a + 5d = 26 …(ii)
Subtracting equation (ii) from (i), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 2
Substituting d = 5 in equation (ii), we get
a + 5(5) = 26
∴ a + 25 = 26
∴ a = 26 – 25 = 1
t1 = a = 1
t2 = t1 + d = 1 + 5 = 6
t3 = t2 + d = 6 + 5 = 11
t4 = t3 + d = 11 + 5 = 16
The required A.P. is 1,6,11,16,….

Question 4.
The A.P. in which 4th term is -15 and 9th term is -30. Find the sum of the first 10 numbers.
Solution:
t4 = -15, t9 = – 30 …[Given]
Since, tn = a + (n – 1)d
∴ t4 = a + (4 – 1)d
∴ – 15 = a + 3d
i. e. a + 3d = -15 …(i)
Also, t9 = a + (9 – 1)d
∴ -30 = a + 8d
i.e. a + 8d = -30 …(ii)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 3
∴ The sum of the first 10 numbers is -195.

Question 5.
Two given A.P.’s are 9, 7, 5, … and 24, 21, 18, … If nth term of both the .progressions are equal then find the value of n and n,h term.
Solution:
The first A.P. is 9, 7, 5,…
Here, a = 9, d = 7- 9 = -2
∴nth term = a + (n – 1)d
= 9 + (n – 1) (-2)
= 9 – 2n + 2
= 11 – 2n
The second A.P. is 24, 21, 18, …
Here, a = 24, d = 21 – 24 = – 3
∴ nth term = a + (n – 1)d
= 24 + (n – 1) (-3)
= 24 – 3n + 3
= 27 – 3n
Since, the nth terms of the two A.P.’s are equal.
∴ 11 – 2n = 27 – 3n
∴ 3n – 2n = 27 – 11
∴ n = 16
∴ t16 = 9 + (16 – 1) (-2)
= 9 + 15 × (-2)
= 9 – 30
∴ t16 = -21
∴ The values of n and nth term are 16 and -21 respectively.

Question 6.
If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is -3, then find the 10th term.
Solution:
for an A.P., let a be the first term and d be the common difference.
According to the first condition,
t3 + tg = 7
∴ a + (3 – 1) d + a + (8 – 1)d = 7 …[∵ tn = a + (n – 1)d]
∴ a + 2d + a + 7d = 7
∴ 2a + 9d = 7 …(i)
According to the second condition,
t7 + t14 = -3
∴ a + (7 – 1)d + a + (14 – 1 )d = -3
∴ a + 6d + a + 13d = -3
∴ 2a + 19 d = – 3 …(ii)
Subtracting equation (i) from (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 4

Question 7.
In an A.P. the first term is -5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
Solution:
Let the number of terms in the A.P. be n and the common difference be d.
Then, a = -5, tn = 45, Sn = 120
Since, tn = a + (n – 1)d
∴ 45 = -5 + (n – 1)d
∴ 45 + 5 = (n – 1)d
∴ (n – 1)d = 50 …(i)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 5
Substituting n = 6 in equation (i), we get
(6 – 1)d = 50
∴ 5d = 50
∴ d = \(\frac { 50 }{ 5 } \) = 10
∴ There are 6 terms in the A.P. and the common difference is 10.

Alternate Method:
Let the number of terms in the A.P. be n.
Then, t1 = a = -5, tn = 45, Sn = 120
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 6
∴ There are 6 terms in the A.P. and the common difference is 10.

Question 8.
Sum of 1 to n natural numbers is 36, then find the value of n.
Solution:
The natural numbers from 1 to n are
1,2, 3, ……, n.
The above sequence is an A.P.
∴ a = 1, d = 2 – 1 = 1
Sn = 36 …[Given]
Now, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ 36 = \(\frac { n }{ 2 } \) [2(1) + (n – 1)(1)]
∴ 36 = \(\frac { n }{ 2 } \) (2 + n – 1)
∴ 36 × 2 = n (n + 1)
∴ 72 = n (n + 1)
∴ 72 = n2 + n
∴ n2 + n – 72 = 0
∴ n2 + 9n – 8n – 72 = 0
∴ n(n + 9) – 8 (n + 9) = 0
∴ (n + 9) (n – 8) = 0
∴ n + 9 = 0 or n – 8 = 0
∴ n = -9 or n = 8
But, n cannot be negative.
∴ n = 8
∴ The value of n is 8.

Question 9.
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
Solution:
Let the three parts of 207 that are in A.P. be
a – d, a, a + d
According to the first condition,
(a – d) + a + (a + d) = 207
∴ 3a = 207
∴ a = \(\frac { 207 }{ 3 } \)
∴ a = 69 …(i)
According to the second condition,
(a – d) × a = 4623
∴ (69 – d) × 69 = 4623 …[From (i)]
∴ 69 – d = \(\frac { 4623 }{ 69 } \)
∴ d = 69 – 67
∴ d = 2
∴ a – d = 69 – 2 = 67
a = 69
a + d = 69 + 2 = 71
∴ The three parts of 207 that are in A.P. are 67, 69 and 71.

Question 10.
There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
Solution:
Since, there are 37 terms in the A.P.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 7
Substituting d = 4 in equation (i), we get
3a + 54(4) = 225
∴ 3a + 216 = 225
∴ 3a = 225 – 216
∴ 3a = 9
∴ a = \(\frac { 9 }{ 3 } \) = 3
∴The required A. P. is
a, a + d, a + 2d, a + 3d, …., a + (n – 1)d
i.e. 3, 3 + 4,3 + 2 × 4, 3 + 3 × 4,…, 3 + (37 – 1)4
i.e. 3, 7,11,15, …,147

Question 11.
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 8
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 9

Question 12.
If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero, (p ≠ q)
Solution:
For an A.P., let a be the first term and d be the common difference.
The sum of first n terms of an A.P. is given by
Sn = [2a + (n – 1)d]
According to the given condition,
Sp = Sq
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 10
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 11
∴ The sum of the first (p + q) terms is zero

Question 13.
If m times the mth term of an A.P. is equal to n times nth term, then show that the (m + n)th term of the A.P. is zero.
Solution:
According to the given condition,
mtm = ntn
∴ m[a + (m – 1)d] = n[a + (n – 1)d]
∴ ma + md(m – 1) = na + nd(n- 1)
∴ ma + m2d – md = na + n2d – nd
∴ ma + m2d – md – na – n2d + nd = 0
∴ (ma – na) + (m2d – n2d) – (md – nd) = 0
∴ a(m – n) + d(m2 – n2) – d(m – n) = 0
∴ a(m – n) + d(m + n) (m – n) – d(m – n) = 0
∴ (m – n)[a + (m + n – 1) d] = 0
∴ [a+ (m + n – 1)d] = 0 …[Dividing both sides by (m – n)]
∴ t(m+n) = 0
∴ The (m + n)th term of the A.P. is zero.

Question 14.
₹ 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 12
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 13

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2

Maharashtra State Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2

Question 1.
Choose the correct answers for the following questions.

i. Which one is the quadratic equation?
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 1
Answer:
(B)

ii. Out of the following equations which one is not a quadratic equation?
(A) x2 + 4x = 11 + x2
(B) x = 4x
(C) 5x2 = 90
(D) 2x – x2 = x2 + 5
Answer:
(A)

iii. The roots of x2 + kx + k = 0 are real and equal, find k.
(A) 0
(B) 4
(C) 0 or 4
(D) 2
Answer:
(C)

iv. For √2 x2 – 5x + √2 = 0, find the value of the discriminant.
(A) -5
(B) 17
(C) √2
(D) 2 √2 – 5
Answer:
(B)

v. Which of the following quadratic equations has roots 3,5?
(A) x2 – 15x + 8 = 0
(B) x2 – 8x + 15 = 0
(C) x2 + 3x + 5 = 0
(D) x2 + 8x – 15 = 0
Answer:
(B)

vi. Out of the following equations, find the equation having the sum of its roots -5.
(A) 3x2 – 15x + 3 = 0
(B) x2 – 5x + 3 = 0
(C) x2 + 3x – 5 = 0
(D) 3x2 + 15x + 3 = 0
Answer:
(D)

vii. √5m2 – √5 m + √5 =0 which of the following statement is true for this given equation?
(A) Real and unequal roots
(B) Real and equal roots
(C) Roots are not real
(D) Three roots
Answer:
(C)

viii. One of the roots of equation x2 + mx – 5 = 0 is 2; find m.
(A) -2
(B) – \(\frac { 1 }{ 2 } \)
(C) \(\frac { 1 }{ 2 } \)
(D) 2
Answer:
(C)

Question 2.
Which of the following equations is quadratic
i. x2 + 2x + 11 = 0
ii. x2 – 2x + 5 = x2
iii. (x + 2)2 = 2x2
Solution:
i. The given equation is
x2 + 2x + 11 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 2, c = 11 are real numbers and
a ≠ 0.
The given equation is a quadratic equation.

ii. The given equation is
x2 – 2x + 5 = x2
∴ x2 – x2 + 2x – 5 = 0
∴ 2x – 5 = 0
Here, x is the only variable and maximum index of the variable is not 2.
∴ The given equation is not a quadratic equation.

iii. The given equation is
(x + 2)2 = 2x2
∴ x2 + 4x + 4 = 2x2
∴ 2x2 – x2 – 4x – 4 = 0
∴ x2 – 4x – 4 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = -4, c = —4 are real numbers and
a ≠ 0.
∴ The given equation is a quadratic equation.

Question 3.
Find the value of discriminant for each of the following equations.
i. 2y2 – y + 2 = 0
ii. 5m2 – m = 0
iii. √5 x2 – x – √5 = 0
Solution:
i. 2y2 – y + 2 = 0
Comparing the above equation with
ay2 + by + c = 0, we get
a = 2, b = -1, c = 2
∴ b2 – 4ac = (-1)2 – 4 × 2 × 2
= 1 – 16
∴ b2 – 4ac = -15

ii. 5m2 – m = 0
∴ 5m2 – m + 0 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 5, b = -1, c = 0
∴ b2 – 4ac = (-1)2 – 4 × 5 × 0
= 1 – 0
∴ b2 – 4ac = 1

iii. √5x2 – x – √5 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = √5, b = -1, c = -√5
∴ b2 – 4ac = (-1)2 – 4 × √5 × √5
= 1 + 20
∴ b2 – 4ac = 21

Question 4.
One of the roots of quadratic equation 2x2 + kx – 2 = 0 is – 2, find k.
Solution:
-2 is one of the roots of the equation
2x2 + kx – 2 = 0.
∴ Putting x = – 2 in the given equation, we get
2(-2)2 + k(-2) -2 = 0
∴ 8 – 2k – 2 = 0
∴ 6 – 2k = 0
∴ 2k = 6
∴ k = \(\frac { 6 }{ 2 } \)
∴ k = 3

Question 5.
Two roots of quadratic equations are given; frame the equation.
i. 10 and -10
ii. 1 – 3√5 and 1 + 3√5
iii. 0 and 7
Solution:
i. Let α = 10 and β = -10
∴ α + β = 10 – 10 = 0
and α × p = 10 × -10 = -100
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 0x + (-100) = 0
∴ x2 – 100 = 0

ii. Let α = 1 – 3 √5 and β = 1 + 3 √5
α + β = 1 – 3 √5 + 1 + 3 √5 = 2
and α × β = (1 – 3√5) (1 + 3 √5)
= (1)2 – (3√5)2
= 1 – 45
= -44
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 2x – 44 = 0

iii. Let α = 0 and β = 7
∴ α + β = 0 + 7 = 7
and α × β = 0 × 7 = 0
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 7x + 0 = 0
∴ x2 – 7x = 0

Question 6.
Determine the nature of roots for each of the quadratic equation.
i. 3x2 – 5x + 7 = 0
ii. √3 x2 + √2 x – 2 √3 = 0
iii. m2 – 2m + 1 = 0
Solution:
i. 3x2 – 5x + 7 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 3, b = -5, c = 7
∴ ∆ = b2 – 4ac
= (-5)2 -4 × 3 × 7
= 25 – 84
∴ ∆ = -59
∴ ∆ < 0
∴ Roots of the given quadratic equation are not real.

ii. √3 x2 + √2 x – 2 √3 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = √3 , b = √2, c = -2√3
∴ ∆ = b2 – 4ac
= (√2)2 – 4 × √3 × (-2√3)
= 2 + 24
∴ ∆ = 26
∴ ∆ > 0
∴ Roots of the given quadratic equation are real and unequal.

iii. m2 – 2m + 1 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 1, b = -2, c = 1
∴ ∆ = b2 – 4ac
= (-2)2 – 4 × 1 × 1
= 4 – 4
∴ ∆ = 0
∴ Roots of the given quadratic equation are real and equal

Question 7.
Solve the following quadratic equations.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 2
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 3
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 4

ii. x2 – \(\frac { 3x }{ 10 } \) – \(\frac { 1 }{ 10 } \) = 0
∴ 10x2 – 3x – 1 = 0
…[Multiplying both sides by 10]
∴ 10x2 – 5x + 2x – 1 = 0
∴ 5x(2x – 1) + 1(2x – 1) = 0
∴ (2x – 1)(5x + 1) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴2x – 1 = 0 or 5x + 1 = 0
∴2x = 1 or 5x = -1
∴ x = –\(\frac { 1 }{ 2 } \) or x = \(\frac { -1 }{ 5 } \)
∴ The roots of the given quadratic equation are \(\frac { 1 }{ 2 } \) and \(\frac { -1 }{ 5 } \)

iii. (2x + 3)2 = 25
∴ (2x + 3)2 – 25 = 0
∴ (2x + 3)2 – (5)2 = 0
∴ (2x + 3 – 5) (2x + 3 + 5) = 0 ….. [∵ a2 – b2 = (a – b) (a + b)]
∴ (2x – 2) (2x + 8) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 2x – 2 = 0 or 2x + 8 = 0
∴ 2x = 2 or 2x = -8
∴ x = \(\frac { 2 }{ 2 } \) or x = \(\frac { -8 }{ 2 } \)
∴ x = 1 or x = -4
∴ The roots of the given quadratic equation are 1 and -4.

iv. m2 + 5m + 5 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 1, b = 5, c = 5
∴ b2 – 4ac = (5)2 – 4 × 1 × 5
= 25 – 20 = 5
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 5

v. 5m2 + 2m+1 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 5, b = 2, c = 1
∴ b2 – 4ac = (2)2 -4 × 5 × 1
= 4 – 20
= -16
∴ b2 – 4ac < 0
∴ Roots of the given quadratic equation are not real.

vi. x2 – 4x – 3 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = -4, c = -3
∴ b2 – 4ac = (-4)2 – 4 × 1 × -3
= 16 + 12
= 28
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 6

Question 8.
Find m, if (m – 12) x2 + 2(m – 12) x + 2 = 0 has real and equal roots.
Solution:
(m – 12) x2 + 2(m – 12)x + 2 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = m – 12, b = 2(m – 12), c = 2
∴ ∆ = b2 – 4ac
= [2(m -12)]2 – 4 × (m – 12) × 2
= 4(m – 12)2 – 8(m – 12)
= 4(m – 12) (m – 12 – 2)
∴ ∆ = 4(m – 12) (m – 14)
Since, the roots are real and equal.
∴ ∆ = 0
∴ 4(m – 12) (m – 14) = 0 (m – 12) (m – 14) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m – 12 = 0 or m – 14 = 0
∴ m = 12 or m = 14
But ,if m = 12, then quadratic coefficient becomes zero.
∴ m ≠ 12
∴m = 14

Question 9.
The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.
Solution:
Let α and β be the roots of the quadratic equation.
According to the given conditions,
α + β = 5 and α3 + β3 = 35
Now, (α + β)3 = α3 + 3α2β + 3αβ2 + β3
∴ (α + β)3 = α3 + β3 + 3αβ (α + β)
∴ (5)3 = 35 + 3αβ(5)
∴ 125 = 35 + 15αβ
∴ 125 – 35 = 15αβ
∴ 15αβ = 90
∴ αβ = \(\frac { 90 }{ 15 } \)
∴ αβ = 6
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 5x + 6 = 0

Question 10.
Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation
2x2 + 2(p + q)x + p2 + q2 = 0.
Solution:
The given quadratic equation is
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 7
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 8
According to the given condition,
Roots of the required quadratic equation are
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 9

Question 11.
Mukund possesses ₹ 50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.
Solution:
Let the amount Sagar possesses be ₹ x.
∴ the amount Mukund possesses = ₹ (x + 50)
According to the given condition,
x(x +50)= 15000
∴ x2 + 50x – 15000 = 0
∴ x2 + 150x- 100x- 15000 = 0
∴ x(x + 150) – 100(x + 150) = 0
∴ (x + 150)(x – 100) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 150 = 0 or x – 100 = 0
∴ x = -150 or x = 100
But, amount cannot be negative.
∴ x= 100 and x + 50 = 100 + 50 = 150
∴ The amount possessed by Sagar and Mukund are ₹ 100 and ₹150 respectively.

Question 12.
The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.
Solution:
Let the numbers be x and y (x > y).
According to the given condition,
x2 – y2 = 120 …(i)
y2 = 2x …(ii)
Substituting y2 = 2x in equation (i), we get
x2 – 2x = 120
∴ x2 – 2x – 120 = 0
∴ x2 – 12x + 10x – 120 = 0
∴ x(x – 12) + 10(x – 12) = 0
∴ (x – 12)(x + 10) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 12 = 0 or x + 10 = 0
∴ x = 12 or x = -10
But x ≠ -10
as, y2 = 2x = 2(-10) = -20 …[Since, the square of number cannot be negative]
∴ x = 12
Smaller number = y2 =2x
∴ y2 = 2 × 12
∴ y2 = 24
∴ y = ± √24 …[Taking square root of both sides]
∴ The smaller number is √24 and greater number is 12 or the smaller number is – √24 and greater number is 12.

Question 13.
Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.
Solution:
Let the number of students be x.
Total number of oranges = 540
∴ the number of oranges each student gets = \(\frac { 540 }{ x } \)
If there were 30 more students, the total number of students = (x + 30) and the total number of oranges each student gets
= (\(\frac { 540 }{ x+30 } \)
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 10Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 11
∴ 30 × 540 = 3x2 + 90 x
∴ 3x2 + 90x= 16200
∴ x2 + 30x – 5400 = 0
…[Dividing both sides by 3]
∴ x2 + 90x – 60x – 5400 = 0
∴ x(x + 90) – 60(x + 90) = 0
∴ (x + 90) (x – 60) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 90 = 0 or x – 60 = 0
∴ x = – 90 or x = 60
But, number of students cannot be negative,
x = 60
∴ The total number of students is 60.

Question 14.
Mr. Dinesh owns an rectangular agricultural farm at village Talvel. The length of the farm is 10 metre more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is \(\frac { 1 }{ 3 } \) of the breadth of the farm. The
area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and side of the pond.
Solution:
Let the breadth of the rectangular farm be x m.
∴ Length of rectangular farm = (2x + 10) m
Area of rectangular farm = Length × Breadth
= (2x + 10) × x
= (2x2+ 10x) sq. m
Now ,side of square shaped pond = \(\frac { x }{ 3 } \) m
∴ Area of square shaped pond = (side)2
= (\(\frac { x }{ 3 } \))2 m
= \(\frac { { x }^{ 2 } }{ 9 } \) m
According to the given condition,
Area of rectangular farm = 20 × Area of pond
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 11
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x = 0 or x – 45 = 0
x = 0 or x = 45
But, breadth of the rectangular farm cannot be zero,
∴ x = 45
Length of rectangular farm
= 2x + 10 = 2(45) + 10 = 100 m
Side of the pond = \(\frac { x }{ 3 } \) = \(\frac { 45 }{ 3 } \) = 15 m
∴ Length and breadth of the farm and the side of the pond are 100 m, 45 m and 15 m respectively.

Question 15.
A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?
Solution:
Let the larger tap take x hours to fill the tank completely.
∴ Part of tank filled by the larger tap in 1 hour = \(\frac { 1 }{ x } \)
Also, the smaller tap takes (x + 3) hours to fill the tank completely.
∴ Part of tank filled by the smaller tap in 1 hour = \(\frac { 1 }{ x+3 } \)
∴Part of tank filled by both the taps in 1 hour
= (\(\frac { 1 }{ x } \) + \(\frac { 1 }{ x+3 } \))
But, the tank gets filled in 2 hours by both the taps.
∴ Part of tank filled by both the taps in 1 hour = \(\frac { 1 }{ 2 } \)
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 12
∴ 2(2x + 3) = x(x + 3)
∴ 4x + 6 = x2 + 3x
∴ x2 + 3x – 4x – 6 = 0
∴ x2 – x – 6 = 0
∴ x2 – 3x + 2x – 6 = 0
∴ x(x – 3) + 2(x – 3) = 0
∴ (x – 3)(x + 2) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 3 = 0 or x + 2 = 0
∴ x = 3 or x = -2
But, time cannot be negative.
∴ x = 3 and x + 3 = 3 + 3 = 6
∴ The larger tap takes 3 hours and the smaller tap takes 6 hours to fill the tank completely.

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4

Maharashtra State Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4

Question 1.
Market value of a share is ₹ 200. If the brokerage rate is 0.3% then find the purchase value of the share.
Solution:
Here, MV = ₹ 200, Brokerage = 0.3%
Brokerage = 0.3% of MV
= \(\frac { 0.3 }{ 100 } \) × 200
= ₹ 0.6
∴ Purchase value of the share = MV + Brokerage
= 200 + 0.6
= ₹ 200.60
∴ Purchase value of the share is ₹ 200.60.

Question 2.
A share is sold for the market value of ₹ 1000. Brokerage is paid at the rate of 0.1%. What is the amount received after the sale?
Solution:
Here, MV = ₹ 1000, Brokerage = 0.1%
∴ Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 1000
∴ Brokerage = ₹ 1
∴ Selling value of the share = MV – Brokerage
= 1000 – 1
= ₹ 999
∴ Amount received after the sale is ₹ 999.

Question 3.
Fill in the blanks given in the contract note of sale-purchase of shares.
(B – buy S – sell)
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 1
Solution:
For buying shares:
Here, Number of shares = 100,
MV of one share = ₹ 45
∴ Total value = 100 × 45
= ₹ 4500
Brokerage= 0.2% of total value 0.2
= \(\frac { 0.2 }{ 100 } \) × 4500
CGST = 9% of brokerage
= \(\frac { 9 }{ 100 } \) × 9 = ₹ 0.81
But, SGST = CGST
∴ SGST = ₹ 0.81
∴ Purchase value of shares
= Total value + Brokerage
= 4500 + 9 + 0.81 + 0.81
= ₹ 4510.62

ii. For selling shares:
Here, Number of shares = 75,
MV of one share = ₹ 200
∴ Total value = 75 × 200
= ₹ 15000
Brokerage = 0.2% of total value
= \(\frac { 0.2 }{ 100 } \) × 15000
= ₹ 30
CGST = 9% of brokerage
= \(\frac { 9 }{ 100 } \) × 30 = ₹ 2.70
But, SGST = CGST
∴ SGST = ₹ 2.70
∴ Selling value of shares = Total value – (Brokerage + CGST + SGST)
= 15000 – (30 + 2.70 + 2.70)
= 15000 – 35.40
= ₹ 14964.60
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 2

Question 4.
Smt. Desai sold shares of face value ₹ 100 when the market value was ₹ 50 and received ₹ 4988.20. She paid brokerage 0.2% and GST on brokerage 18%, then how many shares did she sell?
Solution:
Here, face value of share = ₹ 100,
MV = ₹ 50,
Selling price of shares = ₹ 4988.20,
Rate of brokerage = 0.2%, Rate of GST = 18%
Brokerage = 0.2% of MV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 3

Question 5.
Mr. D’souza purchased 200 shares of FV ₹ 50 at a premium of ₹ 100. He received 50% dividend on the shares. After receiving the dividend he sold 100 shares at a discount of ₹ 10 and remaining shares were sold at a premium of ₹ 75. For each trade he paid the brokerage of ₹ 20. Find whether Mr. D’souza gained or incurred a loss? By how much?
Solution:
For purchasing shares:
Here, FV = ₹ 50, Number of shares = 200,
premium = ₹ 100
MV of 1 share = FV + premium
= 50 + 100
= ₹ 150
∴ MV of 200 shares = 200 × 150 = ₹ 30,000
∴ Mr. D’souza invested amount
= MV of 200 shares + brokerage
= 30,000 + 20
= ₹ 30,020
For selling shares:
Rate of dividend = 50 %, FV = ₹ 50,
brokerage = ₹ 20
Number of shares = 200
Dividend per share = 50% of FV
= \(\frac { 50 }{ 100 } \) × 50
= ₹ 25
∴ Dividend of 200 shares = 200 × 25 = ₹ 5,000
Now, 100 shares are sold at a discount of ₹ 10.
∴ Selling price of 1 share = FV – discount
= 50 – 10
= ₹ 40
∴ Selling price of 100 shares = 100 × 40
= ₹ 4000
∴ Amount obtained by selling 100 shares
= selling price – brokerage
= 4000 – 20
= ₹ 3980
Also, remaining 100 shares are sold at premium of ₹ 75.
∴ selling price of 1 share = FV + premium
= 50 + 75
= ₹ 125
∴ selling price of 100 shares = 100 × 125
= ₹ 12,500
∴ Amount obtained by selling 100 shares
= selling price – brokerage
= 12,500 – 20
= ₹ 12,480
∴ Mr D’souza income = 5000 + 3980 + 12480
= ₹ 21460
Now, Mr D’souza invested amount > income
∴ Mr D’souza incurred a loss.
∴ Loss = amount invested – income
= 30020 – 21460
= ₹ 8560
∴ Mr. D’souza incurred a loss of ₹ 8560.

Question 1.
Nalinitai invested ₹ 6024 in the shares of FV ₹ 10 when the Market Value was ₹ 60. She sold all the shares at MV of ₹ 50 after taking 60% dividend. She paid 0.4% brokerage at each stage of transactions. What was the total gain or loss in this transaction? (Textbook pg. no. 106)
Solution:
Rate of GST is not given in the example, so it is not considered.
For Purchased Shares:
FV = ₹ 10, MV = ₹ 60
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 4

Question 2.
In the above example if GST was paid at 18% on brokerage, then the loss is ₹ 451.92. Verify whether you get the same answer. (Textbook pg, no. 107)
Solution:
For Purchased Shares:
FV = ₹ 10, MV = ₹ 60, sum invested = ₹ 6024, brokerage = 0.4 %, GST = 18%
Brokerage per share = \(\frac { 0.4 }{ 100 } \) × 60 = ₹ 0.24 100
GST per share = \(\frac { 18 }{ 100 } \) × 0.24 = ₹ 0.0432
∴ Cost of one share = 60 + 0.24 + 0.0432
= ₹ 60.2832
∴ Cost of 100 shares = 100 × 60.2832 = ₹ 6028.32
For sold shares:
FV = ₹ 10, MV = ₹ 50, brokerage = 0.4 %,
GST = 18%, Number of shares = 100
Brokerage per share = \(\frac { 0.4 }{ 100 } \) × 50 = ₹ 0.20
GST per share = \(\frac { 18 }{ 100 } \) × 0.20 = ₹ 0.036
Selling price per share = 50 – 0.2 – 0.036
= ₹ 49.764
Selling price of 100 shares = 100 × 49.764
= ₹ 4976.4
Dividend received 60 %
∴ Dividend per share = \(\frac { 60 }{ 100 } \) × 10 = ₹ 6
Dividend on 100 shares = 6 × 100 = ₹ 600
∴ Nalinitai’s income = 4976.4 + 600 = ₹ 5576.4
∴ Cost of 100 shares = ₹ 6028.32
∴ Loss = 6028.32 – 5576.4 = ₹ 451.92
∴ Nalinitai’s loss is ₹ 451.92.

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.4

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.4

Question 1.
On 1st Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.4 1
∴ Sanika’s total saving on 31st December 2016 would be ₹ 70455.

Question 2.
A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 40. Find the amount of the first and last instalment.
Solution:
i. The instalments are in A.P.
Amount repaid in 12 instalments (S12)
= Amount borrowed + total interest
= 8000 + 1360
∴ S12 = 9360
Number of instalments (n) = 12
Each instalment is less than the preceding one by ₹ 40.
∴ d = -40
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.4 2
∴ Amount of the first instalment is ₹ 1000 and that of the last instalment is ₹ 560.

Question 3.
Sachin invested in a national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.
Solution:
i. Amount invested by Sachin in each year are as follows:
5000, 7000, 9000, …
The above sequence is an A.P.
∴ a = 5000, d = 7000 – 5000 = 2000, n = 12

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.4 3
∴ The total amount invested by Sachin in 12 years is ₹ 1,92,000.

Question 4.
There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?
Solution:
i. The number of seats arranged row-wise are as follows:
20, 22, 24,
The above sequence is an A.P.
∴ a = 20, d = 22 – 20 = 2, n = 27

ii. tn = a + (n – 1)d
∴ t15 = 20 + (15 – 1)2
= 20 + 14 × 2
= 20 + 28
∴ t15 = 48
∴ The number of seats in the 15th row is 48.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.4 4
∴ Total seats in the auditorium are 1242.

Question 5.
Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was -30° Celsius then find the temperature on the other five days.
Solution:
Let the temperatures from Monday to Saturday in A.P. be
a, a + d, a + 2d, a + 3d, a + 4d, a + 5d.
According to the first condition,
(a) + (a + 5d) = (a + d) + (a + 5d) + 5°
∴ d = -5°
According to the second condition,
a + 2d = -30°
∴ a + 2(-5°) = -30°
∴ a – 10° = -30°
∴ a = -30° + 10° = -20°
∴ a + d = -20° – 5° = – 25°
a + 3d = -20° + 3(- 5°) = -20° – 15° = -35°
a + 4d = -20° + 4(-5°) = -20° – 20° = -40°
a + 5d = -20° + 5(-5°) = -20° – 25° = -45°
∴ The temperatures on the other five days are
-20°C, -25° C, -35° C, -40° C and -45° C.

Question 6.
On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.
Solution:
i. The number of frees planted row-wise are as follows:
1,2,3,…
The above sequence is an A.P.
∴ a = 1, d = 2 – 1 = 1,n = 25
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.4 5
∴ The total number of trees in 25 rows are 325.

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6

Maharashtra State Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6

Question 1.
Product of Pragati’s age 2 years ago and years hence is 84. Find her present age.
Solution:
Let the present age of Pragati be x years.
∴ 2 years ago,
Age of Pragati = (x – 2) years
After 3 years,
Age of Pragati = (x + 3) years
According to the given condition,
(x – 2) (x + 3) = 84
∴ x(x + 3) – 2(x + 3) = 84
∴ x2 + 3x – 2x – 6 = 84
∴ x2 + x – 6 – 84 = 0
∴ x2 + x – 90 = 0
x2 + 10x – 9x – 90 = 0
∴ x(x + 10) – 9(x + 10) = 0
∴ (x + 10)(x – 9) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 10 = 0 or x – 9 = 0
∴ x = -10 or x = 9
But, age cannot be negative.
∴ x = 9
∴ Present age of Pragati is 9 years.

Question 2.
The sum of squares of two consecutive even natural numbers is 244; find the numbers.
Solution:
Let the first even natural number be x.
∴ the next consecutive even natural number will be (x + 2).
According to the given condition,
x2 + (x + 2)2 = 244
∴ x2 + x2 + 4x + 4 = 244
∴ 2x2 + 4x + 4 – 244 = 0
∴ 2x2 + 4x – 240 = 0
∴ x2 + 2x – 120 = 0 …[Dividing both sides by 2]
∴ x2 + 12x – 10x – 120 = 0
∴ x(x + 12) – 10 (x + 12) = 0
∴ (x + 12) (x – 10) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 12 = 0 or x – 10 = 0
∴ x = -12 or x = 10
But, natural number cannot be negative.
∴ x = 10 and x + 2 = 10 + 2 = 12
∴ The two consecutive even natural numbers are 10 and 12.

Question 3.
In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 1
Solution:
i. Number of trees in a column is x.
ii. Number of trees in a row = x + 5
iii. Total number of trees = x x (x + 5)
iv. According to the given condition,
x(x + 5) = 150
∴ x2 + 5x = 150
∴ x2 + 5x – 150 = 0
v. x2 + 15x – 10x – 150 = 0
∴ x(x+ 15) – 10(x + 15) = 0
∴ (x + 15)(x – 10) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 15 = 0 or x – 10 = 0
∴ x = -15 or x = 10
But, number of trees cannot be negative.
∴ x = 10
vi. Number of trees in a column is 10.
vii. Number of trees in a row = x + 5 = 10 + 5 = 15
∴ Number of trees in a row is 15.

Question 4.
Vivek is older than Kishor by 5 years. The Find their present ages is \(\frac { 1 }{ 6 } \) Find their Present ages
Solution:
Let the present age of Kishor be x.
∴ Present age of Vivek = (x + 5) years
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 2
∴ 6(2x + 5) = x(x + 5)
∴ 12x + 30 = x2 + 5x
∴ x2 + 5x – 12x – 30 = 0
∴ x2 – 7x – 30 = 0
∴ x2 – 10x + 3x – 30 = 0
∴ x(x – 10) + 3(x – 10) = 0
∴ (x – 10)(x + 3) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 10 = 0 or x + 3 = 0
∴ x = 10 or x = – 3
But, age cannot be negative.
∴ x = 10 andx + 5 = 10 + 5 = 15
∴ Present ages of Kishor and Vivek are 10 years and 15 years respectively.

Question 5.
Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.
Solution:
Let the score of Suyash in the first test be x.
∴ Score in the second test = x + 10 According to the given condition,
5(x + 10) = x2
∴ 5x + 50 = x2
∴ x2 – 5x – 50 = 0
∴ x2 – 10x + 5x – 50 = 0
∴ x(x – 10) + 5(x – 10) = 0
∴ (x – 10) (x + 5) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 10 = 0 or x + 5 = 0
∴ x = 10 or x = – 5
But, score cannot be negative.
∴ x = 10
∴ The score of Suyash in the first test is 10.

Question 6.
‘Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is ₹ 40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is ₹ 600, find production cost of one pot and number of pots he makes per day.
Solution:
Let Mr. Kasam make x number of pots on daily basis.
Production cost of each pot = ₹ (10x + 40)
According to the given condition,
x(10x + 40) = 600
∴ 10x2 + 40x = 600
∴ 10x2 + 40x- 600 = 0
∴ x2 + 4x – 60 = 0 …[Dividing both sides by 10]
∴ x2 + 10x – 6x – 60 = 0
∴ x(x + 10) – 6(x + 10) = 0
∴ (x + 10) (x – 6) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 10 = 0 or x – 6 = 0
∴ x = – 10 or x = 6
But, number of pots cannot be negative.
∴ x = 6
∴ Production cost of each pot = 7(10 x + 40)
= ₹ [(10×6)+ 40]
= ₹(60 + 40) = ₹ 100
Production cost of one pot is ₹ 100 and the number of pots Mr. Kasam makes per day is 6.

Question 7.
Pratik takes 8 hours to travel 36 km downstream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current.
Solution:
Let the speed of water current be x km/hr. Speed of boat is 12 km/hr. (x < 12)
In upstream, speed of the water current decreases the speed of the boat and it is the opposite in downstream.
∴ speed of the boat in upstream = (12 – x) km/hr and speed of the boat in downstream = (12 + x) km/hr.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 3
∴ The speed of water current is 6 km/hr.

Question 8.
Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone.
Solution:
Let Nishu take x days to complete the work alone.
∴ Total work done by Nishu in 1 day = \(\frac { 1 }{ x } \)
Also, Pintu takes (x + 6) days to complete the work alone.
∴ Total work done by Pintu in 1 day = \(\frac { 1 }{ x+6 } \)
∴ Total work done by both in 1 day = (\(\frac { 1 }{ x } \) + \(\frac { 1 }{ x+6 } \))
But, both take 4 days to complete the work together.
∴ Total work done by both in 1 day = \(\frac { 1 }{ 4 } \)
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 4
∴ 4(2x + 6) = x(x + 6)
∴ 8x + 24 = x2 + 6x
∴ x2 + 6x – 8x – 24 = 0
∴ x2 – 2x – 24 = 0
∴ x2 – 6x + 4x – 24 = 0
∴ x(x – 6)+ 4(x – 6) = 0
∴ (x – 6) (x + 4) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 6 = 0 or x + 4 = 0
∴ x = 6 or x = -4
But, number of days cannot be negative,
∴ x = 6 and x + 6 = 6 + 6 = 12
∴ Number of days taken by Nishu and Pintu to complete the work alone is 6 days and 12 days respectively.

Question 9.
If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and divisor.
Solution:
Let the natural number be x.
∴ Divisor = x, Quotient = 5x + 6
Also, Dividend = 460 and Remainder = 1
Dividend = Divisor × Quotient + Remainder
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 5
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 9 = 0 or 5x + 51 = 0
∴ x = 9 or x = \(\frac { -51 }{ 5 } \)
But, natural number cannot be negative,
∴ x = 9
∴ Quotient = 5x + 6 = 5(9) + 6 = 45 + 6 = 51
∴ Quotient is 51 and Divisor is 9.

Question 10.
In the given fig. []ABCD is a trapezium, AB || CD and its area is 33 cm2. From the information given in the figure find the lengths of all sides of the []ABCD. Fill in the empty boxes to get the solution.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 6
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 7
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Practice Set 2.6 8

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3

Question 1.
First term and common difference of an A.P. are 6 and 3 respectively; find S27.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 1

Question 2.
Find the sum of first 123 even natural numbers.
Solution:
The even natural numbers are 2, 4, 6, 8,…
The above sequence is an A.P.
∴ a = 2, d = 4 – 2 = 2, n = 123
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 2
∴ The sum of first 123 even natural numbers is 15252.

Question 3.
Find the sum of all even numbers between 1 and 350.
Solution:
The even numbers between 1 and 350 are 2, 4, 6,…, 348.
The above sequence is an A.P.
∴ a = 2, d = 4 – 2 = 2, tn = 348
Since, tn = a + (n – 1)d
∴ 348 = 2 + (n – 1)2
∴ 348 – 2 = (n – 1)2
∴ 346 = (n – 1)2
∴ n – 1 = \(\frac { 346 }{ 2 } \)
∴ n – 1 = 173
∴ n = 173 + 1 = 174
Now, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ S174 = \(\frac { 174 }{ 2 } \) [2 (2) + (174 – 1)2]
= 87(4 + 173 × 2)
= 87(4 + 346)
= 87 × 350
∴ S174 = 30450
∴ The sum of all even numbers between 1 and 350 is 30450.

Question 4.
In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
Solution:
For an A.P., let a be the first term and d be the common difference.
t19 = 52, t38 = 128 …[Given]
Since, tn = a + (n – 1)d
∴ t19 = a + (19 – 1)d
∴ 52 = a + 18d
i.e. a + 18d = 52 …(i)
Also, t38 = a + (38 – 1)d
∴ 128 = a + 37d
i.e. a + 37d = 128 …(ii)
Adding equations (i) and (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 3

Question 5.
Complete the following activity to find the sum of natural numbers between 1 to 140 which are divisible by 4.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 4

Question 6.
Sum of first 55 terms in an A.P. is 3300, find its 28th term.
Solution:
For an A.P., let a be the first term and d be the common difference.
S55 =3300 …[Given]
Since, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ S55 = \(\frac { 55 }{ 2 } \) [2a + (55 – 1)d]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 5

Question 7.
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. (Assume that three consecutive terms in A.P. are a – d, a, a + d.)
Solution:
Let the three consecutive terms in an A.P. be
a – d, a and a + d.
According to the first condition,
a – d + a + a + d = 27
∴ 3a = 27
∴ a = \(\frac { 27 }{ 3 } \)
∴ a = 9 ….(i)
According to the second condition,
(a – d) a (a + d) = 504
∴ a(a2 – d2) = 504
∴ 9(a2 – d2) = 504 …[From (i)]
∴ 9(81 – d2) = 504
∴ 81 – d2 = \(\frac { 504 }{ 9 } \)
∴ 81 – d2 = 56
∴ d2 = 81 – 56
∴ d2 = 25
Taking square root of both sides, we get
d = ± 5
When d = 5 and a =9,
a – d 9 – 5 = 4
a = 9
a + d 9 + 5 = 14
When d = -5 and a = 9,
a – d = 9 – (-5) = 9 + 5 = 14
a = 9
a + d = 9 – 5 = 4
∴ The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.

Question 8.
Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14. (Assume the four consecutive terms in A.P. are a – d, a, a + d, a + 2d.)
Solution:
Let the four consecutive terms in an A.P. be
a – d, a, a + d and a + 2d.
According to the first condition,
a – d + a + a + d + a + 2d = 12
∴ 4a + 2d =12
∴ 2(2a + d) = 12
∴ 2a + d = \(\frac { 12 }{ 2 } \)
∴ 2a + d = 6 …(i)
According to the second condition,
a + d + a + 2d = 14
∴ 2a + 3d = 14 …(ii)
Subtracting equation (i) from (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 6
∴ The four consecutive terms are -3,1,5 and 9.

Question 9.
If the 9th term of an A.P. is zero, then show that the 29th term is twice the 19th term.
To prove: t29 = 2t19
Proof:
For an A.P., let a be the first term and d be the common difference.
t9 = 0 …[Given]
Since, tn = a + (n – 1)d
∴ t9 = a + (9 – 1)d
∴ 0 = a + 8d
∴ a = -8d …(i)
Also, t19 = a + (19 – 1)d
= a + 18d
= -8d + 18d … [From (i)]
∴ t19 = 10d …(ii)
and t29 = a + (29 – 1)d
= a + 28d
= -8d + 28d …[From (i)]
∴ t29 = 20d = 2(10d)
∴ t29 = 2(t19) … [From (ii)]
∴ The 29th term is twice the 19th term.

Question 1.
Find the sum of all odd numbers from 1 to 150. (Textbook pg, no. 71)
Solution:
Odd numbers from 1 to 150 are 1,3, 5, 7,…, 149
Here, difference between any two consecutive terms is 2.
∴ It is an A.P.
∴ a = 1, d = 2
Let us find how many odd numbers are there from 1 to 150, i.e. find the value of n if
tn = 149
tn = a + (n – 1)d
∴ 149 = 1 + (n – 1)2
∴ 149 – 1 = (n – 1)2
∴ \(\frac { 148 }{ 2 } \) = n – 1
∴ 74 = n – 1
∴ n = 74 + 1 = 75

ii. Now, let’s find the sum of 75 numbers
i. e. 1 + 3 + 5 + 7 + … + 149
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 7

Maharashtra Board Class 10 Maths Solutions